# not a homework question

Discussion in 'C Programming' started by Three Headed Monkey, Mar 12, 2008.

1. ### Three Headed MonkeyGuest

write a program in "C" language that computes 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

I tried

#include <stdio.h>

int pow(int n)
{
int i,power;
power=n;
for(i=0;i<n;i=i+1)
power=power*power;
return power;
}

void main()
{
int result;
char ignore;
result= pow(9,pow(8,pow(7,pow(6,pow(5,pow(4,pow(3,pow(2,1))))))));
printf("\nresult is %d", result);
printf("\nPress ENTER");
gets(&ignore);
}

but it does not work.

how to do that in "C" standard language?

I am using lcc-win32 compiler & windows 98.

help!

Three Headed Monkey, Mar 12, 2008

2. ### Ian CollinsGuest

Three Headed Monkey wrote:
> write a program in "C" language that computes 9^(8^(7^(6^(5^(4^(3^(2^1)))))))
>

I don't think you can in standard C, the result will be huge...

--
Ian Collins.

Ian Collins, Mar 12, 2008

3. ### Brice RebsamenGuest

On Mar 12, 12:57 pm, Three Headed Monkey
<> wrote:
> write a program in "C" language that computes 9^(8^(7^(6^(5^(4^(3^(2^1)))))))
>
> I tried
>
> #include <stdio.h>
>
> int pow(int n)
> {
> int i,power;
> power=n;
> for(i=0;i<n;i=i+1)
> power=power*power;
> return power;
>
> }
>
> void main()
> {
> int result;
> char ignore;
> result= pow(9,pow(8,pow(7,pow(6,pow(5,pow(4,pow(3,pow(2,1))))))));
> printf("\nresult is %d", result);
> printf("\nPress ENTER");
> gets(&ignore);
>
> }
>
> but it does not work.
>
> how to do that in "C" standard language?
>
> I am using lcc-win32 compiler & windows 98.
>
> help!

First, you may want to have the pow function take 2 arguments: int
pow(int a, int n) { ... }
Second, the pow function defined in math.h does the job for you,
except that it deals with doubles: float pow(double a, double n)
returns a to the power of n as a float. So if you want to deal with
integers you have to convert the result.
Finally you may want to use a loop to do this. It'd look like this:

#include <math.h>
#include <stdio.h>
int main(){
double res=1;
int n;
for( n=2; n<=9; n++ )
res = pow((double)n,res);
printf("res=%f\n",res);
return 0;
}

Don't forget to link with the math library when compiling (-lm)

However this might overflow, resulting in res reaching inf. You can
try using long double and powl... Or more complicated stuff. Any idea
of what the resulting number might be?

Brice Rebsamen, Mar 12, 2008
4. ### CBFalconerGuest

Three Headed Monkey wrote:
>
> write a program in "C" language that computes
> 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

Since 2 xor 1 is identically zero, that value suffices.

x = 0;

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>

--
Posted via a free Usenet account from http://www.teranews.com

CBFalconer, Mar 12, 2008
5. ### Peter NilssonGuest

Three Headed Monkey <> wrote:
> Subject: not a homework question

Your subject should reflect the nature of your problem,
not merely that you have one.

It's not homework questions we mind, rather it's homework
questions that are quoted verbatim without so much as an
attempt.

> write a program in "C" language that computes
> 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

#include <stdio.h>

int main(void)
{
printf("%d\n", 9^(8^(7^(6^(5^(4^(3^(2^1))))))) );
return 0;
}

If you mean exponentiation, then realise it's pretty big!
Just 5^(4^(3^(2^1))) alone yields a 183000+ digit number.
Raise 6 to the power of that and you have... big!

> I am using lcc-win32 compiler & windows 98.

Ah well... the non-standard extension qfloat should knock
that over easily.

--
Peter

Peter Nilsson, Mar 12, 2008
6. ### user923005Guest

On Mar 11, 9:57 pm, Three Headed Monkey <>
wrote:
> write a program in "C" language that computes 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

I guess that even LCC's qfloat data type will be too small.
Do you have any idea how many digits there are in that number?

user923005, Mar 12, 2008
7. ### user923005Guest

On Mar 11, 10:35 pm, Peter Nilsson <> wrote:
> Three Headed Monkey <> wrote:
>
> > Subject: not a homework question

>
> Your subject should reflect the nature of your problem,
> not merely that you have one.
>
> It's not homework questions we mind, rather it's homework
> questions that are quoted verbatim without so much as an
> attempt.
>
> > write a program in "C" language that computes
> > 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

>
>   #include <stdio.h>
>
>   int main(void)
>   {
>     printf("%d\n", 9^(8^(7^(6^(5^(4^(3^(2^1))))))) );
>     return 0;
>   }
>
> If you mean exponentiation, then realise it's pretty big!
> Just 5^(4^(3^(2^1))) alone yields a 183000+ digit number.
> Raise 6 to the power of that and you have... big!
>
> > I am using lcc-win32 compiler & windows 98.

>
> Ah well... the non-standard extension qfloat should knock
> that over easily.

I doubt it.
Maple with precision set to 10,000,000 digits overflowed:
> Digits=10000000; y := 9.^(8.^(7.^(6.^(5.^(4.^(3.^(2.^1.)))))));
>

10 = 10000000

Error, (in evalf/power) argument too large
>

user923005, Mar 12, 2008
8. ### Richard HeathfieldGuest

Three Headed Monkey said:

>
> write a program in "C" language that computes
> 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

#include <stdio.h>

int main(void)
{
printf("%d\n",
9^(8^(7^(6^(5^(4^(3^(2^1))))))));
return 0;
}

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
"Usenet is a strange place" - dmr 29 July 1999

Richard Heathfield, Mar 12, 2008
9. ### Robert GambleGuest

On Mar 12, 12:57 am, Three Headed Monkey
<> wrote:
> write a program in "C" language that computes 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

Assuming that the caret represents exponentiation (which based on the
context of the rest of your post it is obvious that it does, except
perhaps to Mr. Heathfield ), this isn't a programming problem, it
is a math comprehension problem. Even if it were possible to
calculate the result of this expression you wouldn't be able to store
it. There are more digits in the result than there are particles in
the universe by an unimaginable factor.

--
Robert Gamble

Robert Gamble, Mar 12, 2008
10. ### Micah CowanGuest

Three Headed Monkey <> writes:

> write a program in "C" language that computes 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

No one's mentioned this, so perhaps I'd better:

Are you certain that the expression above (assuming your subject may
just be a slight fib) is intended to represent exponentiation? The ^
is a real operator in C, and means something rather different (**
would be a better choice to represent exponentiation, as it doesn't
have another, real, meaning in C).

Otherwise, you may be best-suited using the standard library's own
pow() function along with a floating point type (double would make
sense, given that's what pow() deals in). OTOH, if you happen to have
an implementation with <tgmath.h> (and don't care to be portable to
them wot don't), you might opt for long double.

--
Micah J. Cowan
Programmer, musician, typesetting enthusiast, gamer...
http://micah.cowan.name/

Micah Cowan, Mar 12, 2008
11. ### Micah CowanGuest

Micah Cowan <> writes:

> Otherwise, you may be best-suited using the standard library's own
> pow() function along with a floating point type (double would make
> sense, given that's what pow() deals in). OTOH, if you happen to have
> an implementation with <tgmath.h> (and don't care to be portable to
> them wot don't), you might opt for long double.

(Obviously, I wasn't thinking too clearly on the magnitude of this
number...)

--
Micah J. Cowan
Programmer, musician, typesetting enthusiast, gamer...
http://micah.cowan.name/

Micah Cowan, Mar 12, 2008
12. ### Ark KhasinGuest

Robert Gamble wrote:
> On Mar 12, 12:57 am, Three Headed Monkey
> <> wrote:
>> write a program in "C" language that computes 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

>
> Assuming that the caret represents exponentiation (which based on the
> context of the rest of your post it is obvious that it does, except
> perhaps to Mr. Heathfield ), this isn't a programming problem, it
> is a math comprehension problem. Even if it were possible to
> calculate the result of this expression you wouldn't be able to store
> it. There are more digits in the result than there are particles in
> the universe by an unimaginable factor.
>
> --
> Robert Gamble

There are more digits in PI than in the number in question .
I mean, in analogy with actual vs. potential infinities, one could
consider a number computed if he can tell for each x what the x-th digit
of the number is.
-- Ark

Ark Khasin, Mar 12, 2008
13. ### WANG CongGuest

On Wed, 12 Mar 2008 05:57:12 +0100ï¼ŒThree Headed Monkey wroteï¼š

> write a program in "C" language that computes
> 9^(8^(7^(6^(5^(4^(3^(2^1)))))))
>
> I tried
>
> #include <stdio.h>
>
> int pow(int n)

Change the name please. C has a std library function with the same
name and it's prototype is:

double pow(double x, double y);

> {
> int i,power;
> power=n;
> for(i=0;i<n;i=i+1)
> power=power*power;
> return power;
> }
>
> void main()

main() is never void in C.

> {
> int result;
> char ignore;
> result=
> pow(9,pow(8,pow(7,pow(6,pow(5,pow(4,pow(3,pow(2,1))))))));

Your _own_ pow only takes one parameter, why here it has two??

> printf("\nresult is %d", result);

I suspect it overflows, since 'result' is only an int.

> printf("\nPress ENTER");
> gets(&ignore);

gets() is considered harmful, NEVER use it.

> }
>
>
> but it does not work.
>
> how to do that in "C" standard language?
>

I am afraid you can't, the result may be too big. You can choose
a non-standard libary that supports huge number operations,
e.g. gmp.

WANG Cong, Mar 12, 2008
14. ### Keith ThompsonGuest

Micah Cowan <> writes:
> Three Headed Monkey <> writes:
>> write a program in "C" language that computes 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

>
> No one's mentioned this, so perhaps I'd better:
>
> Are you certain that the expression above (assuming your subject may
> just be a slight fib) is intended to represent exponentiation? The ^
> is a real operator in C, and means something rather different (**
> would be a better choice to represent exponentiation, as it doesn't
> have another, real, meaning in C).
>
> Otherwise, you may be best-suited using the standard library's own
> pow() function along with a floating point type (double would make
> sense, given that's what pow() deals in). OTOH, if you happen to have
> an implementation with <tgmath.h> (and don't care to be portable to
> them wot don't), you might opt for long double.

There's no need to use <tgmath.h>; just use powl() (which, like
<tgmath.h>, is new in C99, but is perhaps more likely to be supported
as an extension by pre-C99 implementations).

--
Keith Thompson (The_Other_Keith) <>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

Keith Thompson, Mar 12, 2008
15. ### BartcGuest

"Three Headed Monkey" <> wrote in message
news:12984088\$sWSEgdgrr\$...
>
> write a program in "C" language that computes
> 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

So if it's not homework, where does the problem come from? I suspect you
don't really to know the answer to this.

The algorithm can be done neatly in C using integer arithmetic. You were on
the right lines with your code, but you should have tested with a smaller N.

However, it is likely to overflow above N=4. Using bigger ints will help a
little but
will not come near N=9. This is my version tested to N=4:

#include <stdio.h>

int solve(int);

int main(void)
{int n,result;

n=4;

result=solve(n);

printf("Answer for N = %d is %d\n",n,result);

}

int solve(int n)
{
int i,x,a;

if (n<=1) return 1;

x=solve(n-1);

a=n;
for (i=1; i<x; ++i) a*=n;

return a;

}

--
Bart

Bartc, Mar 12, 2008
16. ### Doug MillerGuest

In article <12984088\$sWSEgdgrr\$>, Three Headed Monkey <> wrote:
>
>write a program in "C" language that computes 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

Do you have even the slightest idea how large that number is? Did you perhaps
mean to type * instead of ^ ?
>
>I tried
>
>#include <stdio.h>
>
>int pow(int n)
>{
> int i,power;
> power=n;
> for(i=0;i<n;i=i+1)
> power=power*power;
> return power;
>}

Examine what you've written here a little more carefully, why don't you, and
see exactly what this function calculates. If you want to compute a^b, one
might suppose that you'd probably want a function that accepts both a and b as
input parameters.
>
>void main()
>{
> int result;
> char ignore;
> result= pow(9,pow(8,pow(7,pow(6,pow(5,pow(4,pow(3,pow(2,1))))))));
> printf("\nresult is %d", result);
> printf("\nPress ENTER");
> gets(&ignore);
>}
>
>
>but it does not work.

A little more explanation would be helpful. Start with a complete description
of the manner in which it "does not work", including what you expect it to do
that it does not, and what it does that you expect it not to do.

Doug Miller, Mar 12, 2008
17. ### NoobGuest

Three Headed Monkey wrote:

> write a program in "C" language that computes 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

(I know your post was written tongue-in-cheek, but it's an interesting
problem nonetheless.)

Let u1 = 1 and u(n) = n ^ u(n-1)

Assume base 10.

u1 = 1
u2 = 2
u3 = 9
u4 = 262144
u5 = a number with 183231 digits
u6 = a number with (roughly) 10^183231 digits

u9 might be larger than one googolplex.

http://en.wikipedia.org/wiki/Googolplex

Noob, Mar 12, 2008
18. ### Barry SchwarzGuest

On Tue, 11 Mar 2008 22:25:01 -0700 (PDT), Brice Rebsamen
<> wrote:

>On Mar 12, 12:57 pm, Three Headed Monkey
><> wrote:
>> write a program in "C" language that computes 9^(8^(7^(6^(5^(4^(3^(2^1)))))))
>>
>> I tried
>>
>> #include <stdio.h>
>>
>> int pow(int n)
>> {
>> int i,power;
>> power=n;
>> for(i=0;i<n;i=i+1)
>> power=power*power;
>> return power;
>>
>> }
>>
>> void main()
>> {
>> int result;
>> char ignore;
>> result= pow(9,pow(8,pow(7,pow(6,pow(5,pow(4,pow(3,pow(2,1))))))));
>> printf("\nresult is %d", result);
>> printf("\nPress ENTER");
>> gets(&ignore);
>>
>> }
>>
>> but it does not work.
>>
>> how to do that in "C" standard language?
>>
>> I am using lcc-win32 compiler & windows 98.
>>
>> help!

>
>First, you may want to have the pow function take 2 arguments: int
>pow(int a, int n) { ... }
>Second, the pow function defined in math.h does the job for you,
>except that it deals with doubles: float pow(double a, double n)
>returns a to the power of n as a float. So if you want to deal with

Close. pow returns a double, not a float.

>integers you have to convert the result.
>Finally you may want to use a loop to do this. It'd look like this:
>
>#include <math.h>
>#include <stdio.h>
>int main(){
> double res=1;
> int n;
> for( n=2; n<=9; n++ )
> res = pow((double)n,res);

The cast is superfluous.

> printf("res=%f\n",res);

%g might be better.

> return 0;
>}
>
>Don't forget to link with the math library when compiling (-lm)

A system specific issue that may or may not be applicable to the OP.

>
>However this might overflow, resulting in res reaching inf. You can
>try using long double and powl... Or more complicated stuff. Any idea
>of what the resulting number might be?

Remove del for email

Barry Schwarz, Mar 12, 2008
19. ### Micah CowanGuest

Keith Thompson <> writes:

> Micah Cowan <> writes:
>> Three Headed Monkey <> writes:
>>> write a program in "C" language that computes 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

>>
>> No one's mentioned this, so perhaps I'd better:
>>
>> Are you certain that the expression above (assuming your subject may
>> just be a slight fib) is intended to represent exponentiation? The ^
>> is a real operator in C, and means something rather different (**
>> would be a better choice to represent exponentiation, as it doesn't
>> have another, real, meaning in C).
>>
>> Otherwise, you may be best-suited using the standard library's own
>> pow() function along with a floating point type (double would make
>> sense, given that's what pow() deals in). OTOH, if you happen to have
>> an implementation with <tgmath.h> (and don't care to be portable to
>> them wot don't), you might opt for long double.

>
> There's no need to use <tgmath.h>; just use powl() (which, like
> <tgmath.h>, is new in C99, but is perhaps more likely to be supported
> as an extension by pre-C99 implementations).

Doh! Totally forgot about that.

--
Micah J. Cowan
Programmer, musician, typesetting enthusiast, gamer...
http://micah.cowan.name/

Micah Cowan, Mar 12, 2008
20. ### Keith ThompsonGuest

Noob <root@localhost> writes:
> Three Headed Monkey wrote:
>
>> write a program in "C" language that computes 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

>
> (I know your post was written tongue-in-cheek, but it's an interesting
> problem nonetheless.)
>
> Let u1 = 1 and u(n) = n ^ u(n-1)
>
> Assume base 10.
>
> u1 = 1
> u2 = 2
> u3 = 9
> u4 = 262144
> u5 = a number with 183231 digits
> u6 = a number with (roughly) 10^183231 digits
>
> u9 might be larger than one googolplex.
>
> http://en.wikipedia.org/wiki/Googolplex

(Assuming, of course, that "^" denotes exponentiation.)

u5 is substantially larger than one googol (10^100); it has 183231
digits compared to just 101 digits for one googol.

u6 is substantially larger than one goolplex; it has 10^183231 digits,
compared to just 10^100+1 digits for one googolplex.

u7, u8, and u9 are Really Really Big (but still tiny compared to the
largest numbers that have actually been used in mathematics).

--
Keith Thompson (The_Other_Keith) <>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

Keith Thompson, Mar 13, 2008